MATH 115 SOLUTION SET 4
ANSWERS TO SELECTED PROBLEMS
§
2.1:
1. a) True, b) False, c) False (look at
T
:
R
→
R
deﬁned by
T
(
x
) =
x
+ 1; but
also note that this is not linear.) d) True, e) False (it should be dim
V
there), f)
False (
T
could take everything to zero, what a catastrophe!) g) True, h) False,
because
{
x
1
,x
2
}
might not be linearly independent.
10. Since (2
,
3) =

(1
,
0) + 3(1
,
1), we have
T
(2
,
3) =

T
(1
,
0) + 3
T
(1
,
1) =

(1
,
4) + 3(2
,
5) = (5
,
11).
T
is indeed onetoone.
11. Since (1
,
1) and (2
,
3) are linearly independent, there exists a linear trans
formation
T
:
R
2
→
R
3
that sends these two vectors to any two vectors in
R
3
.
Since (8
,
11) = 2(1
,
1) + 3(2
,
3), apply
T
to both sides to get
T
(8
,
11) = 2(1
,
0
,
2) +
3(1
,

1
,
4) = (5
,

3
,
16).
14. a) Assume
T
is onetoone. Let
S
be a linearly independent subset of
V
;
we will show
T
(
S
) is linearly independent. If
a
1
T
(
v
1
) +
···
+
a
n
T
(
v
n
) = 0 with
v
i
∈
V
and scalars
a
i
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 Spring '10
 SHALOM
 Math, Linear Algebra, Algebra, Vector Space, linearly independent subset

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